(x³+x²y+xy²+y³)(x-y)

=x(x³+x²y+xy²+y³)-y(x³+x²y+xy²+y³)

=x⁴+x³y+x²y²+xy³-x³y-x²y²+xy³+y⁴

= x⁴+y⁴

đề sai bạn xem lại đề

\(x^2y+xy^2+x+y=240\)

\(\Leftrightarrow xy\left(x+y\right)+x+y=240\)

\(\Leftrightarrow11\left(x+y\right)+x+y=240\)

\(\Rightarrow12\left(x+y\right)=240\)

\(\Rightarrow x+y=20\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=20^3-3.11.20=\)

Với x ≥ 0; y ≥ 0 thì x + y ≥ 0

Ta có: x3 + y3 ≥ x2y + xy2

⇔ (x3 + y3) – (x2y + xy2) ≥ 0

⇔ (x + y)(x2 – xy + y2) – xy(x + y) ≥ 0

⇔ (x + y)(x2 – xy + y2 – xy) ≥ 0

⇔ (x + y)(x2 – 2xy + y2) ≥ 0

⇔ (x + y)(x – y)2 ≥ 0 (Luôn đúng vì x + y ≥ 0 ; (x – y)2 ≥ 0)

Dấu « = » xảy ra khi (x – y)2 = 0 ⇔ x = y.

Ta có: \(\left(x^3-x^2y+xy^2-y^3\right)\left(x+y\right)\)

\(=\left[x^2\left(x-y\right)+y^2\left(x-y\right)\right]\left(x+y\right)\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)

\(=x^4-y^4=2^4-\left(\dfrac{1}{2}\right)^4=16-\dfrac{1}{16}=\dfrac{255}{16}\)

Đáp án: A

\(a,=3x^3y^3-3x^2y^3+3x^2y^4+3xy^5\\ b,=\left(2x^3-6x^2+10x-3x^2+9x-15\right):\left(x^2-3x+5\right)\\ =\left[2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)\right]:\left(x^2-3x+5\right)\\ =2x-3\\ c,=\left[x^2\left(x-3\right)+\left(x-3\right)\right]:\left(x-3\right)=x^2+1\)

\(C=xyz+\left(xy+yz+xz\right)+x+y+z-1\)

Ta có ĐT tương đương

\(C=xyz+\left(xy+yz+xz\right)+x+y+z-1=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)

Thay \(x=9\) ; \(y=10\) ; \(z=11\) vào BT có :

\(\left(9-1\right)\left(10-1\right)\left(11-1\right)=720\)

Vậy .........

C = xyz - xy - yz - xz + x + y +z- 1

= xy(z-1) - y(z-1) - x(z-1) + 1(z-1)

(xy-y-x+1)(z-1)

D   =   ( x 3   +   y 3 )   –   x y ( x   +   y )     =   ( x   +   y ) ( x 2   –   x y   +   y 2 )   –   x y ( x   +   y )     =   ( x   +   y ) ( x 2   –   x y   +   y 2   –   x y )     =   ( x   +   y ) [ x ( x   –   y )   –   y ( x   –   y ) ]     =   ( x   +   y ) ( x   –   y ) 2

Vì x = y ó x – y = 0 nên D   =   ( x   +   y ) ( x   –   y ) 2   =   0

Đáp án cần chọn là: D